Started using MatLab a couple of weeks ago, I don't know much proper syntax / terminology.
I'm trying to use a value in a 3x1 matrix as a multiplier in an equation later.
This is to draw a circle with radius and centre point defined by values input by the user.
I have a pop-up window, the values are input by the user and stored in this 3x1 cell (labelled answer).
How do I use the second value of that matrix, answer(2), in the following equation:
x = 'answer(2)' * cos(theta) + xCentre;
This error message appears:
Error using .*
Matrix dimensions must agree.
Error in Disks (line 40)
x = 'answer(2)'.* cos(theta) + xCentre;
In MATLAB, apostrophes ('') define a string. If the name of your matrix is answer, you can refer to its second value by the command answer(2) as mentioned by #Schorsch. For more infos on vectors and matrices, you can check this site.
In addition to what the previous answer says, its important to understand what exactly you are doing before you do it. Only add the ('') if you are defining a string, which generally occurs when dealing with variables. In your case, you simply have a matrix, which is not a string, but rather a set of numbers. You can simply do answer(2) as aforementioned, because answer(2) calls up the second value in your matrix while 'answer(2)' has you trying to define some variable that does not exist.
the most important thing is truly understanding what you are doing to avoid the basic syntax errors.
Related
I'm trying to vectorize one function in Matlab, but I have a problem with assigning values.
function [val] = clenshaw(coeffs,x)
b=zeros(1,length(coeffs)+2);
for k=length(coeffs):-1:2
b(k)=coeffs(k)-b(k+2)+2*b(k+1).*x;
end
val=coeffs(1)-b(3)+b(2).*x;
The purpose of this function is to use Clenshaw's algorithm to compute a value of one polynomial with coefficients "coeffs" at point x.
It work fine when x is a single value, but I'd like it to work with vector of arguments too.
When I try to pass a vector I get an error:
Unable to perform assignment because the left
and right sides have a different number of
elements.
Error in clenshaw (line 7)
b(k)=coeffs(k)-b(k+2)+2*b(k+1).*x;
I understand that there is a problem, because I'm trying to assign vector to a scalar variable b(k).
I tried making b a matrix instead of a vector, however I still cannot get the return output I'd like to have which would be a vector of values of this function at points from vector x.
Thank you for helping and sorry if something isn't entirely clear, because English is not my native language.
The vectorized version of your function looks like this:
function [val] = clenshaw(coeffs,x)
b=zeros(length(x),length(coeffs)+2);
for k=length(coeffs):-1:2
b(:,k)=coeffs(k)-b(:,k+2)+2*b(:,k+1).*transpose(x);
end
val=coeffs(1)-b(:,3)+b(:,2).*transpose(x);
end
b needs to be a matrix. In your loop, you have to perform every operation per row of b. So you need to write b(:,k) instead of b(k). Since b(:,k) is a vector and not a scalar, you also have to be careful with the dimensions when using the .* operator. To get the correct results, you need to transpose x. The same goes for the calculation of val. If you don't like the transposition, just swap the rows and cols of b and you get this:
function [val] = clenshaw(coeffs,x)
b=zeros(length(coeffs)+2, length(x));
for k=length(coeffs):-1:2
b(k,:)=coeffs(k)-b(k+2,:)+2*b(k+1,:).*x;
end
val=coeffs(1)-b(3,:)+b(2,:).*x;
end
However, the first version returns a column vector and the second a row vector. So you might need to transpose the result if the vector type is important.
Hi, I am trying to write a function as per the question. I have tried to create four sub-matrices which are the reverse of each other and then multiply to give the products demanded by the question. My attempt:
function T = custom_blocksT(n,m)
T(1:end,end-1:1);
T(1:end,end:-1:1)*2;
T(1:end,end:-1:1)*3;
T(1:end,end:-1:1)*4;
What I'm unsure of is
(i) What do the the indivual sub-matrices(T(1:end,end-1:1);)need to be equal to? I was thinking of(1:3)?
(ii) I tried to create a generic sub-matrix which can take any size matrix input using end was this correct or can't you do that? I keep getting this error
Undefined function or variable 'T'.
Error in custom_blocksT (line 2)
T(1:end,end-1:1);
I have searched the Matlab documentation and stacked overflow, but the problem is I'm not quite sure what I'm supposed to be looking for in terms of solving this question.
If someone could help me I would be very thankfull.
There are many problems with your function:
function T = custom_blocksT(n,m)
T(1:end,end-1:1);
T(1:end,end:-1:1)*2;
T(1:end,end:-1:1)*3;
T(1:end,end:-1:1)*4;
end
This is an extremely basic question, I highly recommend you find and work through some very basic MATLAB tutorials before continuing, even before reading this answer to be honest.
That said here is what you should have done and a bit of what you did wrong:
First, you are getting the error that T dos not exist because it doesn't. The only variables that exist in your function are those that you create in the function or those that are passed in as parameters. You should have passed in T as a parameter, but instead you passed in n and m which you don't use.
In the question, they call the function using the example:
custom_blocks([1:3;3:-1:1])
So you can see that they are only passing in one variable, your function takes two and that's already a problem. The one variable is the matrix, not it's dimensions. And the matrix they are passing in is [1:3;3:-1:1] which if you type in the command line you will see gives you
[1 2 3
3 2 1]
So for your first line to take in one argument which is that matrix it should rather read
function TOut = custom_blocks(TIn)
Now what they are asking you to do is create a matrix, TOut, which is just different multiples of TIn concatenated.
What you've done with say TIn(1:end,end-1:1)*2; is just ask MATLAB to multiple TIn by 2 (that's the only correct bit) but then do nothing with it. Furthermore, indexing the rows by 1:end will do what you want (i.e. request all the rows) but in MATLAB you can actually just use : for that. Indexing the columns by end-1:1 will also call all the columns, but in reverse order. So in effect you are flipping your matrix left-to-right which I'm sure is not what you wanted. So you could have just written TIn(:,:) but since that's just requesting the entire matrix unchanged you could actually just write TIn.
So now to multiply and concatenate (i.e. stick together) you do this
TOut = [TIn, TIn*2; TIn*3, TIn*4]
The [] is like a concatenate operation where , is for horizontal and ; is for vertical concatenation.
Putting it all together:
function TOut = custom_blocks(TIn)
TOut = [TIn, TIn*2; TIn*3, TIn*4];
end
function[df] = getDiscountFactor(t,T,r,i)
y=getYearFraction(t,T);
a=r(i,1)
df=1/((1+a)^y);
end
This is my code, thing is that y cannot do the (1+a)^y operation because it says it's a vector, but it is only a vector value r(i,1). When I call it in command line it prints a as 0.03 and then it shows this error
a =
0.0300
Error using ^
Inputs must be a scalar and a square
matrix.
To compute elementwise POWER, use
POWER (.^) instead.
If i use .^ operator problem is solved but df gets into a vector and I need it to be a single number.
You've checked the first input to see that it's a scalar, so that presumably means the other input, y, is not a square matrix. If you check what y is too, you will probably see what's gone wrong.
And if not, does matlab have a function that tells you what type matlab thinks an object is? Run that on a and on y. Or to be extra pedantic, run that on 1+a. That should clear things up.
here is the code listing and i got the above mentiond error at line nests(r,c)=nests(r,c)+stepsize.*randn(size(nests(r,c))); please let me now what is wrong with my code as i m new to matlab
for r = 1:numb_of_nest % for each particle
for c = 1:4
u=randn(size(nests(r,c)))*sigma;
v=randn(size(nests(r,c)));
step=u./abs(v).^(1/beta);
nests(r,c)=nests(r,c)+stepsize.*randn(size(nests(r,c)));
% Apply simple bounds/limits
ns_tmp=nests(r,c);
I=ns_tmp<Lb(c);
ns_tmp(I)=Lb(I);
% Apply the upper bounds
J=ns_tmp>Ub(c);
ns_tmp(J)=Ub(J);
% Update this new move
nests(r,c)=ns_tmp;
end
end
This error happens when you assign a value of some dimension m x n to a subscripted variable of different dimension.
In your case, assuming nests has no third dimension, you're assigning to a scalar (1x1) variable. This only works if the value you're trying to assign also is a scalar. Since you get the error, it probably isn't. The only place where your dimensions can be non-scalar is stepsize, so to fix this error, make sure stepsize is a scalar value.
According to the definition you gave in an earlier comment (stepsize=0.01*step.*(nests(r,c)-best);), this problem translates to make sure best is a scalar value. Possibly by subscripting, I can't tell you exactly how since I don't know what best is.
step=u./abs(v).^(1/beta);
nests(r,c)=nests(r,c)+stepsize.*randn(size(nests(r,c)));
Here you're assigning a value to the variable step, but then using a different variable called stepsize that hasn't been assigned a value anywhere in this code. Is this intentional? If not, stepsize is probably some leftover variable from previous code which is messing up the dimensions and giving you this error.
In addition to the above, is nests an ordinary two-dimensional matrix in your code? If so, taking size(nests(r,c)) every time is unnecessary - since you're giving two subscripts, the result is only going to be 1 all the time. Or is nests a cell array perhaps? In that case, you might want to index using curly braces { } instead of ordinary parantheses, to get the size of the matrix that's sitting inside the cell.
I have recently been tasked with using a first derivative filter on an image of myself. The instructor said that I should first fix the value of y and preform f(x+1) - f(x) on the rows and then fix the new "X" values and preform f(y+1)-f(y) on the columns.
Note: I have been asked to do this task manually, not using filter2() or any other programmed function, so please do not suggest that I use filter2() or similar. Thanks!
I tried calling up all the pixels and subtracting each successive one by doing
fid = fopen('image.raw')
myimage = fread(fid,[512 683], '*int8')
fclose(fid)
imsz = size(myimage)
x = imsz(1)
for I = 1:512
for J = 1:683
X(I) - X(I-1) = XX
But it doesnt seem to work, and I dont quite understand why. If you could help me, or point me in the right direction, I would be very appreciative.
First of all, your code is syntatically incorrect:
There is no end statement to any of your loops, and besides, you don't even need loops here.
You seem to read your image into the variable myimage, but you're using an undefined variable X when attempting to calculate the derivative.
The order of your assignment statements is reversed. The variable you wish to assign to should be written in the left hand part of the assignement.
I strongly suggest that you read online tutorials and get yourself familiar with MATLAB basics before taking on more complicated tasks.
As to your specific problem:
MATLAB encourages vectorized operations, i.e operations on entire arrays (vectors or matrices) at once. To subtract adjacent values in an array, what you're basically doing is subtracting two arrays, shifted by one element with respect to each other. For one dimensional arrays, that would translate in MATLAB to:
a(2:end) - a(1:end-1)
where a is your array. The end keyword specifies the last index in the array.
To compute the derivative of an image (a 2-D matrix), you need to decide along which axis you want to perform that operation. To approximate the derivate along the y-axis, do this:
X(2:end, :) - X(1:end-1, :)
You can verify that this gives you the same result as diff(X, 1) (or simply diff(X)). To compute the approximate derivative along the x-axis, which is equivalent to diff(X, 2), do this:
X(:, 2:end) - X(:, 1:end-1)
The colon (:) is the same as writing 1:end as the array subscript for the corresponding dimension.
If your filtered image is div then
for Y = 1:682
for X = 1:511
div(X, Y) = myimage(X + 1, Y + 1) - myimage(X,Y);
end
end
Remember the last row and the last column are not filtered!